Question: $f(x) = 4x^{3}-4x$ $h(x) = -6x^{2}-5x+6+3(f(x))$ $g(x) = 6x^{3}-3x^{2}-x-4(f(x))$ $ f(g(1)) = {?} $
Solution: First, let's solve for the value of the inner function, $g(1)$ . Then we'll know what to plug into the outer function. $g(1) = 6(1^{3})-3(1^{2})-1-4(f(1))$ To solve for the value of $g$ , we need to solve for the value of $f(1)$ $f(1) = 4(1^{3})+(-4)(1)$ $f(1) = 0$ That means $g(1) = 6(1^{3})-3(1^{2})-1+(-4)(0)$ $g(1) = 2$ Now we know that $g(1) = 2$ . Let's solve for $f(g(1))$ , which is $f(2)$ $f(2) = 4(2^{3})+(-4)(2)$ $f(2) = 24$